A 20 mL$20 \text{ mL}$ sample of a sodium iodide solution yields 4.74 g$4.74 \text{ g}$ of silver iodide precipitate when treated with an excess of silver nitrate solution. The molarity of the initial sodium iodide solution is _________ M (as the nearest integer value). Given molar masses: Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol⁻¹$\text{Na} = 23, \, \text{I} = 127, \, \text{Ag} = 108, \, \text{N} = 14, \, \text{O} = 16 \text{ g mol}^{-1}$.

Related Formula

Core Logic

First, calculate the input moles for each reactant:

• Moles of Fe₃O₄ = 2.32 × 10³ × 10³ g232 g mol⁻¹ = 10,000 moles$$Fe_{3}O_{4} = \frac{2.32 \times 10^3 \times 10^3\text{ g}}{232\text{ g mol}^{-1}} = 10,000\text{ moles}$$
• Moles of CO = 2.8 × 10² × 10³ g28 g mol⁻¹ = 10,000 moles$$CO = \frac{2.8 \times 10^2 \times 10^3\text{ g}}{28\text{ g mol}^{-1}} = 10,000\text{ moles}$$

Next, identify the limiting reagent by comparing the available moles to the stoichiometric coefficients:

• For Fe₃O₄$Fe_3O_4$: (10000)/(1) = 10000$\frac{10000}{1} = 10000$
• For CO$CO$: (10000)/(4) = 2500$\frac{10000}{4} = 2500$

Since 2500 < 10000$2500 < 10000$, carbon monoxide (CO$CO$) is the limiting reagent.

Now, determine the production yield of iron based on the limiting reagent (CO$CO$):

Convert these moles into kilograms to find the final mass (x$x$):

Pattern Recognition

Always identify the limiting reagent first by normalizing the mole quantities with their respective stoichiometric coefficients before calculating product yields.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Related Formula

Mass percentage definition:

Density conversion equation:

Core Logic

A value of 75% w/w HNO₃$75\%\text{ w/w HNO}_3$ implies that 75 g$75\mathrm{\ g}$ of pure HNO₃$\text{HNO}_3$ is present in 100 g$100\mathrm{\ g}$ of solution.

We need to find the volume that provides exactly 30 g$30\mathrm{\ g}$ of pure acid solute.

Step 1: Calculate Solution Mass and Volume

Mass of solution needed for 30 g$30\mathrm{\ g}$ solute:

Converting mass to volume using solution density (1.25 g/mL$1.25\mathrm{\ g/mL}$):

Pattern Recognition

Break concentration steps down clearly: Mass of solute arrow Mass of solution arrow Volume of solution$\text{Mass of solute} \rightarrow \text{Mass of solution} \rightarrow \text{Volume of solution}$. Combining operations: Volume = Mass solute% × 100density = (30)/(75) × (100)/(1.25) = 0.4 × 80 = 32$\text{Volume} = \frac{\text{Mass solute}}{\%} \times \frac{100}{\text{density}} = \frac{30}{75} \times \frac{100}{1.25} = 0.4 \times 80 = 32$.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Related Formula

Absolute zero (0 K$0\text{ K}$) represents the lowest theoretical temperature limit.

Core Logic

Analyzing each statement based on foundational definitions :

• (A) & (B) Incorrect: Mass is the actual matter present; weight is the gravitational force exerted on that mass. These definitions are reversed in the statements.
• (C) Correct: Volume correctly defines the space occupied by a substance .
• (D) Correct: Celsius values can be negative, whereas Kelvin scale strictly defaults to absolute zero (0 K$0\text{ K}$) as minimum .
• (E) Correct: Precision measures how close experimental trials lie relative to each other .

Therefore, statements (C), (D), and (E) are correct.

Pattern Recognition

Absolute temperature scale (Kelvin) can never possess real negative values because 0 K$0\text{ K}$ represents complete cessation of molecular motion.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Core Logic

Statement (I): Potassium hydrogen phthalate (KHP) is widely used as a primary standard in analytical chemistry for standardizing strong bases like NaOH$NaOH$. This is because it is highly pure, non-hygroscopic, stable, and has a relatively high molar mass, making its concentration reliable and stable over time.

Statement (II): KHP is a weak acid and NaOH$NaOH$ is a strong base. The titration of a weak acid with a strong base yields an equivalence point in the weakly basic range (pH > 7). Phenolphthalein changes colour in the pH range 8.3 to 10.0, making it the perfect indicator for this titration.

Step 1: Evaluate Statements

Statement I is correct. Statement II is correct.

Pattern Recognition

Weak Acid vs Strong Base arrow$\rightarrow$ Equivalence pH > 7 arrow$\rightarrow$ Phenolphthalein is the indicator of choice.

Chapter Mix

Class 11 Chemistry: Equilibrium Class 11 Chemistry: Some Basic Concepts of Chemistry

Related Formula

Core Logic

From the balanced chemical equation: 3 moles of PbCl₂ react with 2 moles of (NH₄)₃PO₄$3 \text{ moles of } PbCl_2 \text{ react with } 2 \text{ moles of } (NH_4)_3PO_4$.

Let's find the limiting reagent (L.R.) by dividing given millimoles by stoichiometric coefficients: For PbCl₂$PbCl_2$: (72)/(3) = 24$\frac{72}{3} = 24$ For (NH₄)₃PO₄$(NH_4)_3PO_4$: (50)/(2) = 25$\frac{50}{2} = 25$

Since 24 < 25$24 < 25$, PbCl₂$PbCl_2$ is the limiting reagent and will completely consume.

Step 1: Calculate Product Moles

Moles of Pb₃(PO₄)₂$Pb_3(PO_4)_2$ formed depends entirely on PbCl₂$PbCl_2$. 3 mmol of PbCl₂$PbCl_2$ produces 1 mmol of Pb₃(PO₄)₂$Pb_3(PO_4)_2$. Therefore, 72 mmol of PbCl₂$PbCl_2$ will produce: (1)/(3) × 72 = 24 ~mmol$\frac{1}{3} \times 72 = 24 \mathrm{~mmol}$ of Pb₃(PO₄)₂$Pb_3(PO_4)_2$.

Pattern Recognition

Always identify the Limiting Reagent by taking the ratio n / coefficient$n / \text{coefficient}$. The smallest ratio dictates the extent of the reaction.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry